A car is traveling from a rate of 20 m/s and comes to a complete stop. If the car was decelerating at -5 m/s/s, how long did it take for the car to stop?

Astudent made a graph plotting the progress of a reaction over time. the student forgot to label the y-axis of the graph. a graph is shown with two graph lines. one graph line starts at a higher position on the y axis and slopes downwards towards the right. the other graph line starts at a lower position on the y axis and slopes upwards towards the right. the two graph lines stop short of intersecting each other and continue as separate lines which gradually become straight and parallel to the x axis. a vertical line is shown at a point where the two graph lines finally became parallel to the x axis. this vertical line is labeled equilibrium. the title on the x axis is time and an arrow pointing towards the right is shown above time. the title on the y axis is left blank. what best explains the label that the student should use on the y-axis? amount, because as the amount of product decreases, the amount of reactant increases over time. reaction rate, because forward and backward reaction become equal at equilibrium. amount, because the amounts of reactants and products become constant after equilibrium is reached. reaction rate, as the rate of forward reaction increases and rate of backward reaction decreases over time.

For the reaction c6h14(g) & longrightarrow; c6h6(g) + 4h2(g), δp(h2)/δt was found to be 2.5 x 10-2 atm/s, where δp(h2) is the change in pressure of hydrogen. determine δp(c6h14)/δt for this reaction at the same time.

Select the correct value for the indicated bond angle in each of the compounds. o−o−oo−o−o angle of o3 90° 109.5° < 109.5° 120° < 120° 180° f−b−ff−b−f angle of bf3 180° < 109.5° < 120° 120° 109.5° 90° f−o−ff−o−f angle of of2 < 120° 120° 90° 109.5° 180° < 109.5° cl−be−clcl−be−cl angle of becl2 90° 109.5° 180° 120° < 109.5° < 120° f−p−ff−p−f angle of pf3 90° 109.5° < 109.5° 180° 120° < 120° h−c−hh−c−h angle of ch4 90° < 109.5° 180° 120° < 120° 109.5°